Magnetism Question 202
Question: If a magnet of length 10 cm and pole strength 40 A-m is placed at an angle of $45^\circ$ in an uniform induction field of intensity $2 \times 10^4$ T, the couple acting on it is
[Pb. PMT 1999; MH CET (Med.) 1999]
Options:
A) $0.5656 \times 10^4 $ N-m
B) $0.5656 \times 10 ^{-3} $ N-m
C) $0.656 \times 10^{-4} $ N-m
D) $0.656 \times 10^5 $ N-m
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Answer:
Correct Answer: B
Solution:
$ \tau =MB\sin \theta =( mL )B\sin \theta $
$ =(40\times 10\times {{10}^{-2}})\times 2\times {{10}^{-4}}\times \sin 45{}^\circ $
$ =0.565\times {{10}^{-3}}N-m $
BETA
